OFFSET
1,2
LINKS
Jean-Marc Falcoz, Table of n, a(n) for n = 1..40000
EXAMPLE
1 is in the sequence because 12/(1+2) is the integer 4;
2 is in the sequence though 23/(2+3) is not an integer because if we compute floor(23/(2+3)) we get 4, then if we use this 4 to compute floor(34/(3+4)) we get 4, then again floor(44/(4+4)) = 5 and in the end 45/(4+5) is the integer 5;
3 is in the sequence though 34/(3+4) is not an integer, because if we apply the "floor" trick again, we will end in an integer: floor(34/(3+4)) = 4, then floor(44/(4+4)) = 5 and 45/(4+5) is the integer 5;
4 is in the sequence because 45/(4+5) is the integer 5;
5 is not in the sequence because 56/(5+6) is not an integer and even if we repeatedly apply the "floor" trick, we will be stuck in a loop: floor(56/(5+6)) = 5, then floor(65/(6+5)) = 5, then floor(55/(5+5)) = 5, then again floor(55/(5+5)) = 5, etc. So 5 will never produce an integer at the end.
6 is not in the sequence for the same reason: floor(67/(6+7)) = 5, then floor(75/(7+5)) = 6, then floor(56/(5+6)) = 5, then floor(65/(6+5)) = 5, then floor(55/(5+5)) = 5, then again floor(55/(5+5)) = 5, etc. So 6 will never produce an integer at the end.
7 is not in the sequence for the same reason again: floor(78/(7+8)) = 5, then floor(85/(8+5)) = 6, then floor(56/(5+6)) = 5, then floor(65/(6+5)) = 5, then floor(55/(5+5)) = 5, then again floor(55/(5+5)) = 5, etc. So 7 will never produce an integer at the end.
. . .
10 is in the sequence though 1011/(10+11) is not an integer, because if we apply the "floor" trick again, we will end on an integer: floor(1011/(10+11)) = 48, then floor(1148/(11+48)) = 19, then floor(4819/(48+19)) = 71, then floor(1971/(19+71)) = 21, then floor(7121/(71+21)) = 77, then floor(2177/(21+77) = 22, then 7722/(77+22) is the integer 78.
Etc.
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Eric Angelini and Jean-Marc Falcoz, Jul 08 2018
STATUS
approved